Tietze extension theorem

This article gives the statement, and possibly proof, of a basic fact in topology.

Statement

Suppose $X$ is a normal space (i.e., a topological space that is T1 and where any two disjoint closed subsets can be separated by disjoint open subsets). Suppose $A$ is a closed subset of $X$, and $f:A\to [0,1]$ is a continuous map. Then, there exists a continuous map $g:X\to [0,1]$ such that the restriction of $g$ to $A$ is $f$.

Facts used

1. Urysohn's lemma: This states that for, given two closed subsets $A,B$ of a normal space $X$, there is a continuous function $h:X\to [0,1]$ such that $h(A)=0$ and $h(B)=1$.

Proof

Note that since $[0,1]$ is homeomorphic to $[-1,1]$, it suffices to prove the result replacing $[0,1]$ with $[-1,1]$. We will also freely use that any closed interval is homeomorphic to $[0,1]$, so Urysohn's lemma can be stated replacing $[0,1]$ by any closed interval.

Given: A normal space $X$. A closed subset $A$ of $X$. A continuous function $f:A\to [-1,1]$.

To prove: There exists a continuous function $g:X\to [-1,1]$ such that the restriction of $g$ to $A$ is $f$.

Proof: We write $f_{0}=f$.

Step no.	Construction/assertion	Given data used	Previous steps used	Facts used	Explanation
1	Let $\!C_{1}=f^{-1}([-1,-1/3])$ and $\!D_{1}=f^{-1}([1/3,1])$	Existence of $f:A\to [0,1]$	--	--
2	$C_{1}$ and $D_{1}$ are disjoint closed subsets of $A$	$f$ is continuous	(1)	inverse image of closed subset under continuous map is closed	$C_{1}$ and $D_{1}$ are closed subsets of $A$ since they are both the inverse image of a closed set under a continuous map. Moreover, they are disjoint since their images under $f$ are disjoint.
3	$C_{1}$ and $D_{1}$ are disjoint closed subsets of $X$	$A$ is closed in $X$	(2)	(2)	By step (2), $C_{1}$ and $D_{1}$ are closed in $A$, which is closed in $X$. Thus, $C_{1}$ and $D_{1}$ are closed in $X$.
4	Construct a continuous function $\!g_{1}:X\to [-1/3,1/3]$ such that $\!g_{1}(C_{1})=-1/3$ and $\!g_{1}(D_{1})=1/3$		(3)	(1) (Urysohn's lemma)	Urysohn's lemma guarantees a continuous function $h:X\to [0,1]$ such that $h(C_{1})=0$ and $h(D_{1})=1$. Define $g_{1}(x):=(2h(x)-1)/3$.
5	Define $\!f_{1}=f_{0}-g_{1}$ as a function on $A$. Note that $\!f_{1}$ is a continuous function on $\!A$ taking values in $\![-2/3,2/3]$.	($f=f_{0}$ is continuous)	(4)	sum or difference of continuous functions on subsets of a topological space is continuous on their intersection.	$f_{1}$ is continuous since both $f_{0}$ and $g_{1}$ are continuous. For the range, note the following. For $x\in A$, there are three possibilities for the interval in which $f_{0}(x)$ lies: $[-1,-1/3]$, $[-1/3,1/3]$, and $[1/3,1]$. In the first case, $g_{1}(x)=-1/3$, so $f_{0}(x)-g_{1}(x$ is in $[-1-(-1/3),-1/3-(-1/3)]=[-2/3,0]$. In the second case, both $f_{0}(x)$ and $g_{1}(x)$ are in $[-1/3,1/3]$, so by the triangle inequality, the difference is in $[-2/3,2/3]$. In the third case, $g_{1}(x)=1/3$ so $f_{0}(x)-g_{1}(x)$ is in $[1/3-1/3,1-1/3]=[0,2/3]$.
6	We proceed iteratively as follows: from the previous stage, we have a continuous function $f_{i}$ on the closed subset $A$ taking values in $\![-(2/3)^{i},(2/3)^{i}]$.
7	Let $\!C_{i+1}=f_{i}^{-1}[-(2/3)^{i},-(1/3)(2/3)^{i}]$ and $\!D_{i+1}=f_{i}^{-1}[(1/3)(2/3)^{i},(2/3)^{i}]$.
8	$C_{i+1}$ and $D_{i+1}$ are disjoint closed subsets of $A$.		(6): $f_{i}$ is continuous	inverse image of closed subset under continuous map is closed	$C_{1}$ and $D_{1}$ are closed subsets of $A$ since they are both the inverse image of a closed set under a continuous map. Moreover, they are disjoint since their images under $f$ are disjoint.
9	$C_{i+1}$ and $D_{i+1}$ are disjoint closed subsets of $X$	$A$ is closed in $X$	(8)	(2) (closed subsets of closed subsets are closed)	By step (8), $C_{i+1}$ and $D_{i+1}$ are closed in $A$, which is closed in $X$. Thus, $C_{i+1}$ and $D_{i+1}$ are closed in $X$.
10	Find a continuous function $g_{i+1}:X\to [-(1/3)(2/3)^{i},(1(3/(2/3)^{i}]$ such that $\!g_{i+1}(C_{i+1})=(-1/3)(2/3)^{i}$ and $\!g_{i+1}(D_{i+1})=(1/3)(2/3)^{i}$.		(9)	(1) (Urysohn's lemma)	We use fact (1) to get a function to $[0,1]$, then multiply by 2 and subtract 1 to get a function to $[-1,1]$, then scale it by the factor of $(1/3)(2/3)^{i}$.
11	Define $\!f_{i+1}=f_{i}-g_{i+1}$. $f_{i+1}$ is a continuous function on $A$ taking values in $\![-(2/3)^{i+1},(2/3)^{i+1}]$.		(6), (10)		(Same logic as for (5))
12	Define the following function $g:X\to [-1,1]$: $g=\sum _{i=1}^{\infty }g_{i}$. This function is well defined.		(10)		Note that the absolute value of $g_{i}$ is bounded by the geometric progression $(1/3)+(1/3)(2/3)+(1/3)(2/3)^{2}+\dots =1$. Similarly, the lower bound is $-1$. Further, since $g_{n}$ are bounded by the geometric progression in absolute value, the series $g_{n}$ converges. So the sum is well-defined and takes values in $[-1,1]$.
13	The function $g$ is continuous		(10)		This follows from the fact that each $g_{i}$ is continuous and the co-domain of the $g_{i}$s approaches zero. Fill this in later
14	$\!g|_{A}=f$				Let $x\in A$. Then, $\!f_{1}(x)=f(x)-g_{1}(x)$. Inductively, $f_{n}(x)=f(x)-\sum _{i=1}^{n}g_{i}(x)$. Since the upper and lower bound on $f_{n}$ tend to zero as $n\to \infty$, $f_{n}(x)\to 0$ as $n\to \infty$. Thus, $f(x)=\sum _{i=1}^{\infty }g_{i}(x)=g(x)$ for $x\in A$.